Solve r^2=(r-10)^2+20^3

$$r ^ { 2 } = ( r - 10 ) ^ { 2 } + 20 ^ { 3 }$$

Solve for r

$$r=405$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(r-10\right)^{2}.

$$r^{2}=r^{2}-20r+100+20^{3}$$

Calculate 20 to the power of 3 and get 8000.

$$r^{2}=r^{2}-20r+100+8000$$

Add 100 and 8000 to get 8100.

$$r^{2}=r^{2}-20r+8100$$

Subtract r^{2} from both sides.

$$r^{2}-r^{2}=-20r+8100$$

Combine r^{2} and -r^{2} to get 0.

$$0=-20r+8100$$

Swap sides so that all variable terms are on the left hand side.

$$-20r+8100=0$$

Subtract 8100 from both sides. Anything subtracted from zero gives its negation.

$$-20r=-8100$$

Divide both sides by -20.

$$r=\frac{-8100}{-20}$$

Divide -8100 by -20 to get 405.

$$r=405$$